Ccna PPT

2
Communications and Services
Certifications

4
CCNA Exam
Exam Number - 640-801
Total Marks - 1000
Duration – 90 Mts
Passing score – 849
Questions -45-55
Multiple Choice
Simulations
Drag and Drop

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Benefits
Peer Validation
 Personal
 Potential Employer
Career advancement

7
Data Networks
Sharing data through the use of floppy disks is not an efficient
or cost-effective manner.
Businesses needed a solution that would successfully address
the following three problems:
• How to avoid duplication of equipment and resources
• How to communicate efficiently
• How to set up and manage a network
Businesses realized that networking technology could increase
productivity while saving money.

8
Networking Devices
Equipment that connects directly to a network segment is
referred to as a device.
These devices are broken up into two classifications.
 End-user devices
 Network devices
End-user devices include computers, printers, scanners, and
other devices that provide services directly to the user.
Network devices include all the devices that connect the end-
user devices together to allow them to communicate.

9
Network Interface Card
A network interface card (NIC) is a printed circuit board
that provides network communication capabilities to and
from a personal computer. Also called a LAN adapter.

10
Hub
Connects a group of Hosts

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Switch
Switches add more
intelligence to data transfer
management.

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Router
Routers are used to connect networks together
Route packets of data from one network to another
Cisco became the de facto standard of routers because of their high-
quality router products
Routers, by default, break up a broadcast domain

13
Network Topologies
Network topology defines the structure of the network.
One part of the topology definition is the physical topology,
which is the actual layout of the wire or media.
The other part is the logical topology,which defines how the
media is accessed by the hosts for sending data.

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Bus Topology
A bus topology uses a single backbone cable that is
terminated at both ends.
All the hosts connect directly to this backbone.

15
Ring Topology
A ring topology connects one host to the next and the last
host to the first.
This creates a physical ring of cable.

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Star Topology
A star topology connects all cables to a central point of
concentration.

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Extended Star Topology
An extended star topology links individual stars together by
connecting the hubs and/or switches.This topology can extend
the scope and coverage of the network.

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Mesh Topology
A mesh topology is implemented to provide as much
protection as possible from interruption of service.
Each host has its own connections to all other hosts.
 Although the Internet has multiple paths to any one
location, it does not adopt the full mesh topology.

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Physical and Logical Topology

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LANs, MANs, & WANs
One early solution was the creation of local-area network
(LAN) standards which provided an open set of guidelines for
creating network hardware and software, making equipment
from different companies compatible.
What was needed was a way for information to move
efficiently and quickly, not only within a company, but also
from one business to another.
The solution was the creation of metropolitan-area networks
(MANs) and wide-area networks (WANs).

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Virtual Private Network
A VPN is a private network that is constructed within a public network
infrastructure such as the global Internet. Using VPN, a telecommuter
can access the network of the company headquarters through the
Internet by building a secure tunnel between the telecommuter’s PC
and a VPN router in the headquarters.

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What Are The Components Of A
Network ?
Main OfficeBranch Office
Home
Office
Mobile
Users
Internet

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Network Structure &
Hierarchy
Distribution
Layer
Core Layer
Access
Layer

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Institute of Electrical and Electronics
Engineers (IEEE) 802 Standards
 IEEE 802.1: Standards related to network management.
 IEEE 802.2: General standard for the data link layer in the OSI
Reference Model. The IEEE divides this layer into two sublayers --
the logical link control (LLC) layer and the media access control
(MAC) layer.
 IEEE 802.3: Defines the MAC layer for bus networks that use
CSMA/CD. This is the basis of the Ethernet standard.
 IEEE 802.4: Defines the MAC layer for bus networks that use a
token-passing mechanism (token bus networks).
 IEEE 802.5: Defines the MAC layer for token-ring networks.
 IEEE 802.6: Standard for Metropolitan Area Networks (MANs)

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Why do we need the OSI Model?
To address the problem of networks increasing in size and in number, the
International Organization for Standardization (ISO) researched many
network schemes and recognized that there was a need to create a network
model
This would help network builders implement networks that could
communicate and work together
ISO therefore, released the OSI reference model in 1984.

32
Don’t Get Confused.
ISO - International Organization for Standardization
OSI - Open System Interconnection
IOS - Internetwork Operating System
To avoid confusion, some people say “International
Standard Organization.”

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The OSI Reference Model
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
The OSI Model will be
used throughout your
entire networking
career!
Memorize it!

34
OSI Model
Data Flow
Layers
Transport
Data-Link
Network
Physical
Application
(Upper)
Layers
Session
Presentation
Application

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Layer 7 - The Application Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
This layer deal with
networking
applications.
Examples:
 Email
 Web browsers
PDU - User Data
Each of the layers have Protocol Data Unit (PDU)

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Layer 6 - The Presentation Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
This layer is responsible
for presenting the data in
the required format which
may include:
Code Formatting
Encryption
Compression
PDU - Formatted Data

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Layer 5 - The Session Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
This layer establishes, manages, and
terminates sessions between two
communicating hosts.
Creates Virtual Circuit
Coordinates communication between systems
Organize their communication by offering
three different modes
Simplex
Half Duplex
Full Duplex
Example:
 Client Software
( Used for logging in)
PDU - Formatted Data

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Half Duplex
• It uses only one wire pair with a digital signal running in
both directions on the wire.
• It also uses the CSMA/CD protocol to help prevent
collisions and to permit retransmitting if a collision does
occur.
• If a hub is attached to a switch, it must operate in half-
duplex mode because the end stations must be able to
detect collisions.
• Half-duplex Ethernet—typically 10BaseT—is only about
30 to 40 percent efficient because a large 10BaseT
network will usually only give you 3 to 4Mbps—at most.

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Full Duplex
In a network that uses twisted-pair cabling, one pair is used to carry the transmitted
signal from one node to the other node. A separate pair is used for the return or
received signal. It is possible for signals to pass through both pairs simultaneously.
The capability of communication in both directions at once is known as full duplex.

40
Layer 4 - The Transport Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
This layer breaks up the data from the
sending host and then reassembles it in the
receiver.
It also is used to insure reliable data
transport across the network.
Can be reliable or unreliable
Sequencing
Acknowledgment
Retransmission
Flow Control
PDU - Segments

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Layer 3 - The Network Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
Sometimes referred to as the “Cisco Layer”.
End to End Delivery
Provide logical addressing that routers use for
path determination
Segments are encapsulated
Internetwork Communication
Packet forwarding
Packet Filtering
Makes “Best Path Determination”
Fragmentation
PDU – Packets – IP/IPX

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Layer 2 - The Data Link Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
Performs Physical Addressing
This layer provides reliable transit of
data across a physical link.
Combines bits into bytes and
bytes into frames
Access to media using MAC address
Error detection, not correction
LLC and MAC
Logical Link Control performs Link
establishment
MAC Performs Access method
PDU - Frames
Preamble DMAC SMAC Data length DATA FCS

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Layer 1 - The Physical Layer
7 Application
6 Presentation
5 Session
4 Transport
3 Network
2 Data Link
1 Physical
This is the physical media
through which the data,
represented as electronic signals,
is sent from the source host to
the destination host.
Move bits between devices
Encoding
PDU - Bits

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Data Encapsulation
Transport
Data-Link
Physical
Network
Upper-Layer Data
Upper-Layer DataTCP Header
DataIP Header
DataLLC Header
0101110101001000010
DataMAC Header
Presentation
Application
Session
Segment
Packet
Bits
Frame
PDU
FCS
FCS

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OSI Model Analogy
Application Layer - Source Host
After riding your new bicycle a few times in
Bangalore, you decide that you want to give it to
a friend who lives in DADAR, Mumbai.

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OSI Model Analogy
Presentation Layer - Source Host
Make sure you have the proper directions to
disassemble and reassemble the bicycle.

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OSI Model Analogy
Session Layer - Source Host
Call your friend and make sure you have his
correct address.

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OSI Model Analogy
Transport Layer - Source Host
Disassemble the bicycle and put different pieces
in different boxes. The boxes are labeled
“1 of 3”, “2 of 3”, and “3 of 3”.

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OSI Model Analogy
Network Layer - Source Host
Put your friend's complete mailing address (and
yours) on each box.Since the packages are too
big for your mailbox (and since you don’t have
enough stamps) you determine that you need to
go to the post office.

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OSI Model Analogy
Data Link Layer – Source Host
Bangalore post office takes possession of the
boxes.

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OSI Model Analogy
Physical Layer - Media
The boxes are flown from Bangalore to Mumbai.

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OSI Model Analogy
Data Link Layer - Destination
Dadar post office receives your boxes.

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OSI Model Analogy
Network Layer - Destination
Upon examining the destination address,
Dadar post office determines that your
boxes should be delivered to your written
home address.

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OSI Model Analogy
Transport Layer - Destination
Your friend calls you and tells you he got all 3
boxes and he is having another friend named
BOB reassemble the bicycle.

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OSI Model Analogy
Session Layer - Destination
Your friend hangs up because he is done talking
to you.

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OSI Model Analogy
Presentation Layer - Destination
BOB is finished and “presents” the bicycle to
your friend. Another way to say it is that your
friend is finally getting him “present”.

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OSI Model Analogy
Application Layer - Destination
Your friend enjoys riding his new bicycle in
Dadar.

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Data Flow Through a Network

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Type of Transmission
Unicast
Multicast
Broadcast

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Broadcast Domain
A group of devices receiving broadcast frames
initiating from any device within the group
Routers do not forward broadcast frames,
broadcast domains are not forwarded from one
broadcast to another.

63
Collision
 The effect of two nodes sending transmissions
simultaneously in Ethernet. When they meet on the
physical media, the frames from each node collide and
are damaged.

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Collision Domain
The network area in Ethernet over which frames
that have collided will be detected.
Collisions are propagated by hubs and repeaters
Collisions are Not propagated by switches,
routers, or bridges

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Physical Layer
Defines
• Media type
• Connector type
• Signaling type
802.3
Physical802.3 is responsible for LANs based on the carrier sense multiple access
collision detect (CSMA/CD) access methodology. Ethernet is an example
of a CSMA/CD network.

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Physical Layer:
Ethernet/802.3
Hub
Hosts
Host
10Base2—Thin Ethernet
10Base5—Thick Ethernet
10BaseT—Twisted Pair

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Device Used At Layer 1
A B C D
Physical
• All devices are in the same collision domain.
• All devices are in the same broadcast domain.
• Devices share the same bandwidth.

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Hubs & Collision Domains
• More end stations means
more collisions.
• CSMA/CD is used.

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Layer 2
DataSource Address FCSLengthDestination Address
Variable266 4
0000.0C xx.xxxx
Vendor
Assigned
IEEE Assigned
MAC Layer—802.3
Preamble
Ethernet II
uses “Type”
here and
does not use
802.2.
MAC Address
8Number of Bytes
synchronize senders and receivers

70
Devices On Layer 2
(Switches & Bridges)
• Each segment has its own collision domain.
• All segments are in the same broadcast domain.
Data-Link
OR1 2 3 1 24

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Switches
• Each segment is its
own collision domain.
• Broadcasts are
forwarded to all
segments.
Memory
Switch

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Layer 3 : Network Layer
• Defines logical
source and
destination
addresses
associated with a
specific protocol
• Defines paths
through network
Network
IP, IPX
Data-LinkPhysical
EIA/TIA-232
V.35
802.2
802.3

73
Layer 3 : (cont.)
Data
Source
Address
Destination
Address
IP Header
172.15.1.1
NodeNetwork
Logical
Address
Network Layer End-Station Packet
Route determination occurs at this layer, so a packet must include a source and
destination address.
Network-layer addresses have two components: a network component for
internetwork routing, and a node number for a device-specific address. The
example in the figure is an example of an IP packet and address.

74
Layer 3 (cont.)
11111111 11111111 00000000 00000000
10101100 00010000 01111010 11001100
Binary
Mask
Binary
Address
172.16.122.204 255.255.0.0
172 16 122 204
255
Address Mask
255 0 0
Network Host

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Device On Layer 3
Router
• Broadcast control
• Multicast control
• Optimal path
determination
• Traffic management
• Logical addressing
• Connects to WAN
services

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Layer 4 : Transport Layer
• Distinguishes between
upper-layer applications
• Establishes end-to-end
connectivity between
applications
• Defines flow control
• Provides reliable or
unreliable services for
data transfer
Network
IPXIP
Transport
SPXTCP UDP

77
Reliable Service
Synchronize
Acknowledge, Synchronize
Acknowledge
Data Transfer
(Send Segments)
Sender Receiver
Connection Established

78
How They Operate
Hub Bridge Switch Router
Collision Domains:
1 4 4 4
Broadcast Domains:
1 1 1 4

80
Why Another Model?
Although the OSI reference model is universally recognized, the
historical and technical open standard of the Internet is
Transmission Control Protocol / Internet Protocol (TCP/IP).
The TCP/IP reference model and the TCP/IP protocol stack
make data communication possible between any two
computers, anywhere in the world, at nearly the speed of light.
The U.S. Department of Defense (DoD) created the TCP/IP
reference model because it wanted a network that could survive
any conditions, even a nuclear war.

81
TCP/IP Protocol Stack
7
6
5
4
3
2
5
4
3
2
Application
Presentation
Session
Transport
Network
Data-Link
Physical1
Application
Transport
Internet
Data-Link
Physical
1

82
Application Layer Overview
*Used by the Router
Application
Transport
Internet
Data-Link
Physical
File Transfer
- TFTP*
- FTP*
- NFS
E-Mail
- SMTP
Remote Login
- Telnet*
- rlogin*
Network Management
- SNMP*
Name Management
- DNS*

83
Transport Layer Overview
Transmission Control
Protocol (TCP)
User Datagram
Protocol (UDP)
Application
Transport
Internet
Data-Link
Physical
Connection-
Oriented
Connectionless

84
TCP Segment Format
Source Port (16) Destination Port (16)
Sequence Number (32)
Header
Length (4)
Acknowledgment Number (32)
Reserved (6) Code Bits (6) Window (16)
Checksum (16) Urgent (16)
Options (0 or 32 if Any)
Data (Varies)
20
Bytes
Bit 0 Bit 15 Bit 16 Bit 31

85
Port Numbers
TCP
Port
Numbers
F
T
P
Transport
Layer
T
E
L
N
E
T
D
N
S
S
N
M
P
T
F
T
P
S
M
T
P
UDP
Application
Layer
21 23 25 53 69 161
R
I
P
520

86
TCP Port Numbers
Source
Port
Destination
Port
…
Host A
1028 23 …
SP DP
Host Z
Telnet Z
Destination port = 23.
Send packet to my
Telnet
application.

88
Send SYN
(seq = 100 ctl = SYN)
SYN Received
Send SYN, ACK
(seq = 300 ack = 101
ctl = syn,ack)
Established
(seq = 101 ack = 301
ctl = ack)
Host A Host B
1
2
3
SYN Received
TCP Three-Way
Handshake/Open Connection

89
Opening & Closing Connection

90
Windowing
• Windowing in networking means the quantity of data
segments which is measured in bytes that a machine can
transmit/send on the network without receiving an
acknowledgement

91• Window Size = 1
Sender Receiver
Send 1
Receive 1
Receive ACK 2
Send ACK 2
Send 2
Receive 2
Receive ACK 3
Send ACK 3
Send 3
Receive 3
Receive ACK 4 Send ACK 4
TCP Simple Acknowledgment

92
TCP Sequence and
Acknowledgment Numbers
Source
Port
Destination
Port
…
Sequence Acknowledgment
1028 23
Source Dest.
11
Seq.
101
Ack.
1028 23
Source Dest.
10
Seq.
100
Ack.
102823
Source Dest.
11
Seq.
100
Ack.
102823
Source Dest.
12
Seq.
101
Ack.
I just got number
11, now I need
number 12.
I just
sent number
11.

93
Windowing
 There are two window sizes—one set to 1 and one set to
3.
 When you’ve configured a window size of 1, the sending
machine waits for an acknowledgment for each data
segment it transmits before transmitting another
 If you’ve configured a window size of 3, it’s allowed to
transmit three data segments before an
acknowledgment is received.

95
Transport Layer Reliable Delivery

96
Flow Control
 Another function of the transport layer is to provide
optional flow control.
 Flow control is used to ensure that networking devices
don’t send too much information to the destination,
overflowing its receiving buffer space, and causing it to
drop the sent information
 The purpose of flow control is to ensure the destination
doesn't get overrun by too much information sent by the
source

97
Flow Control
SEQ 1024
SEQ 2048
SEQ 3072
A
B
3072
3

98
User Datagram Protocol (UDP)
User Datagram Protocol (UDP) is the connectionless transport protocol in
the TCP/IP protocol stack.
UDP is a simple protocol that exchanges datagrams, without
acknowledgments or guaranteed delivery. Error processing and
retransmission must be handled by higher layer protocols.
UDP is designed for applications that do not need to put sequences of
segments together.
The protocols that use UDP include:
• TFTP (Trivial File Transfer Protocol)
• SNMP (Simple Network Management Protocol)
• DHCP (Dynamic Host Control Protocol)
• DNS (Domain Name System)

99
• No sequence or acknowledgment fields
UDP Segment Format
Source Port (16) Destination Port (16)
Length (16)
Data (if Any)
1Bit 0 Bit 15 Bit 16 Bit 31
Checksum (16)
8
Bytes

101
Internet Layer Overview
• In the OSI reference model, the network layer
corresponds to the TCP/IP Internet layer.
Internet Protocol (IP)
Internet Control Message
Protocol (ICMP)
Address Resolution
Protocol (ARP)
Reverse Address
Resolution Protocol (RARP)
Application
Transport
Internet
Data-Link
Physical

102
IP Datagram
Version
(4)
Destination IP Address (32)
Options (0 or 32 if Any)
Data (Varies if Any)
1Bit 0 Bit 15 Bit 16 Bit 31
Header
Length (4)
Priority &Type
of Service (8)
Total Length (16)
Identification (16)
Flags
(3) Fragment Offset (13)
Time-to-Live (8) Protocol (8) Header Checksum (16)
Source IP Address (32)
20
Bytes

103
•Determines destination upper-layer protocol
Protocol Field
Transport
Layer
Internet
Layer
TCP UDP
Protocol
Numbers
IP
176

104
Internet Control Message
Protocol
Application
Transport
Internet
Data-Link
Physical
Destination
Unreachable
Echo (Ping)
Other
ICMP
1

105
Address Resolution Protocol
• Map IP MAC
• Local ARP
172.16.3.1
IP: 172.16.3.2
Ethernet: 0800.0020.1111
172.16.3.2
IP: 172.16.3.2 = ???
I heard that broadcast.
The message is for me.
Here is my Ethernet
address.
I need the
Ethernet
address of
176.16.3.2.

106
Reverse ARP
• Map MAC IP
Ethernet: 0800.0020.1111
IP: 172.16.3.25
Ethernet: 0800.0020.1111 IP = ???
What is
my IP
address?
I heard that
broadcast.
Your IP
address is
172.16.3.25.

108
Found by Xerox Palo Alto Research Center (PARC) in
1975
Original designed as a 2.94 Mbps system to connect
100 computers on a 1 km cable
Later, Xerox, Intel and DEC drew up a standard
support 10 Mbps – Ethernet II
Basis for the IEEE’s 802.3 specification
Most widely used LAN technology in the world
Origin of Ethernet

109
10 Mbps IEEE Standards - 10BaseT
• 10BaseT  10 Mbps, baseband,
over Twisted-pair cable
• Running Ethernet over twisted-pair
wiring as specified by IEEE 802.3
• Configure in a star pattern
• Twisting the wires reduces EMI
• Fiber Optic has no EMI
Unshielded twisted-pair
RJ-45 Plug and Socket

110
 Unshielded Twisted Pair Cable (UTP)
most popular
maximum length 100 m
prone to noise
Category 1
Category 2
Category 3
Category 4
Category 5
Category 6
Voice transmission of traditional telephone
For data up to 4 Mbps, 4 pairs full-duplex
Twisted Pair Cables

111
 Baseband Transmission
 Entire channel is used to transmit a single digital signal
 Complete bandwidth of the cable is used by a single signal
 The transmission distance is shorter
 The electrical interference is lower
 Broadband Transmission
 Use analog signaling and a range of frequencies
 Continuous signals flow in the form of waves
 Support multiple analog transmission (channels)
Modem Broadband
Transmission
Network
Card
Baseband
Transmission
Baseband VS Broadband

113
Straight-through cable pinout

118
Straight-Thru or Crossover
Use straight-through cables for the following cabling:
 Switch to router
 Switch to PC or server
 Hub to PC or server
Use crossover cables for the following cabling:
 Switch to switch
 Switch to hub
 Hub to hub
 Router to router
 PC to PC
 Router to PC

120
Decimal to Binary
100 = 1
101 = 10
102 = 100
103 = 1000
1
10
100
1000
172 – Base 10
1
2
4
8
16
32
64
128
10101100– Base 2
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
10101100
172
2
70
100
172
0
0
4
8
0
32
0
128
172

121
Base 2 Number System
101102 = (1 x 24 = 16) + (0 x 23 = 0) + (1 x 22 = 4) +
(1 x 21 = 2) + (0 x 20 = 0) = 22

122
Converting Decimal to Binary
Convert 20110 to binary:
201 / 2 = 100 remainder 1
100 / 2 = 50 remainder 0
50 / 2 = 25 remainder 0
25 / 2 = 12 remainder 1
12 / 2 = 6 remainder 0
When the quotient is 0, take all the remainders in
reverse order for your answer: 20110 = 110010012

124
Hex to Binary to Decimal Chart

125
– Unique addressing allows communication
between end stations.
– Path choice is based on destination address.
• Location is represented by an address
Introduction to TCP/IP
Addresses
172.18.0.2
172.18.0.1
172.17.0.2172.17.0.1
172.16.0.2
172.16.0.1
SA DAHDR DATA
10.13.0.0 192.168.1.0
10.13.0.1 192.168.1.1

126
IP Addressing
255 255 255 255
Dotted
Decimal
Maximum
Network Host
128
64
32
16
8
4
2
1
11111111 11111111 11111111 11111111
10101100 00010000 01111010 11001100
Binary
32 Bits
172 16 122 204Example
Decimal
Example
Binary
1 8 9 16 17 24 25 32
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1

127
•Class A:
•Class B:
•Class C:
•Class D: Multicast
•Class E: Research
IP Address Classes
Network Host Host Host
Network Network Host Host
Network Network Network Host
8 Bits 8 Bits 8 Bits 8 Bits

128
IP Address Classes
1
Class A:
Bits:
0NNNNNNN Host Host Host
8 9 16 17 24 25 32
Range (1-126)
1
Class B:
Bits:
10NNNNNN Network Host Host
8 9 16 17 24 25 32
Range (128-191)
1
Class C:
Bits:
110NNNNN Network Network Host
8 9 16 17 24 25 32
Range (192-223)
1
Class D:
Bits:
1110MMMM Multicast Group Multicast Group Multicast Group
8 9 16 17 2425 32
Range (224-239)

129
Host Addresses
172.16.2.2
172.16.3.10
172.16.12.12
10.1.1.1
10.250.8.11
10.180.30.118
E1
172.16 12 12
Network Host
. . Network Interface
172.16.0.0
10.0.0.0
E0
E1
Routing Table
172.16.2.1
10.6.24.2
E0

130
Classless Inter-Domain Routing
(CIDR)
• Basically the method that ISPs (Internet Service
Providers) use to allocate an amount of
addresses to a company, a home
• Ex : 192.168.10.32/28
• The slash notation (/) means how many bits are
turned on (1s)

132
11111111
Determining Available Host
Addresses
172 16 0 0
10101100 00010000 00000000 00000000
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
Network Host
00000000 00000001
11111111 11111111
11111111 11111110
...
...
00000000 00000011
11111101
1
2
3
65534
65535
65536
–
...
2
65534
N
2N – 2 = 216 – 2 = 65534

133
IP Address Classes Exercise
Address Class Network Host
10.2.1.1
128.63.2.100
201.222.5.64
192.6.141.2
130.113.64.16
256.241.201.10

134
IP Address Classes Exercise
Answers
Address Class Network Host
10.2.1.1
128.63.2.100
201.222.5.64
192.6.141.2
130.113.64.16
256.241.201.10
A
B
C
C
B
Nonexistent
10.0.0.0
128.63.0.0
201.222.5.0
192.6.141.0
130.113.0.0
0.2.1.1
0.0.2.100
0.0.0.64
0.0.0.2
0.0.64.16

135
Subnetting
Subnetting is logically dividing the network
by extending the 1’s used in SNM
Advantage
Can divide network in smaller parts
Restrict Broadcast traffic
Security
Simplified Administration

136
Formula
 Number of subnets – 2x-2
Where X = number of bits borrowed
 Number of Hosts – 2y-2
Where y = number of 0’s
 Block Size = Total number of Address
Block Size = 256-Mask

137
Subnetting
 Classful IP Addressing SNM are a set of 255’s and 0’s.
 In Binary it’s contiguous 1’s and 0’s.
 SNM cannot be any value as it won’t follow the rule of
contiguous 1’s and 0’s.
 Possible subnet mask values
– 0
– 128
– 192
– 224
– 240
– 248
– 252
– 254
– 255

138
• Network 172.16.0.0
172.16.0.0
Addressing Without Subnets
172.16.0.1 172.16.0.2 172.16.0.3
…...
172.16.255.253 172.16.255.254

139
• Network 172.16.0.0
Addressing with Subnets
172.16.1.0 172.16.2.0
172.16.3.0
172.16.4.0

140
Subnet Addressing
172.16.2.200
172.16.2.2
172.16.2.160
172.16.2.1
172.16.3.5
172.16.3.100
172.16.3.150
E0
172.16
Network
Network Interface
172.16.0.0
172.16.0.0
E0
E1
New Routing Table
2 160
Host
. .
172.16.3.1
E1

141
Subnet Addressing
172.16.2.200
172.16.2.2
172.16.2.160
172.16.2.1
172.16.3.5
172.16.3.100
172.16.3.150
172.16.3.1
E0
E1
172.16 2 160
Network Host
. . Network Interface
172.16.2.0
172.16.3.0
E0
E1
New Routing Table
Subnet

142
Subnet Mask
172 16 0 0
255 255 0 0
255 255 255 0
IP
Address
Default
Subnet
Mask
8-Bit
Subnet
Mask
Network Host
Network Host
Network Subnet Host
• Also written as “/16,” where 16 represents the number of 1s
in the mask
• Also written as “/24,” where 24 represents the number of
1s in the mask
11111111 11111111 00000000 00000000

143
Decimal Equivalents of Bit
Patterns
0 0 0 0 0 0 0 0 = 0
1 0 0 0 0 0 0 0 = 128
1 1 0 0 0 0 0 0 = 192
1 1 1 0 0 0 0 0 = 224
1 1 1 1 0 0 0 0 = 240
1 1 1 1 1 0 0 0 = 248
1 1 1 1 1 1 0 0 = 252
1 1 1 1 1 1 1 0 = 254
1 1 1 1 1 1 1 1 = 255
128 64 32 16 8 4 2 1

144
16
Network Host
172 0 0
10101100
11111111
10101100
00010000
11111111
00010000
00000000
00000000
10100000
00000000
00000000
•Subnets not in use—the default
00000010
Subnet Mask Without Subnets
172.16.2.160
255.255.0.0
Network
Number

145
•Network number extended by eight bits
Subnet Mask with Subnets
16
Network Host
172.16.2.160
255.255.255.0
172 2 0
10101100
11111111
10101100
00010000
11111111
00010000
11111111
00000010
10100000
00000000
00000000
00000010
Subnet
Network
Number
128
192
224
240
248
252
254
255

146
Subnet Mask with Subnets
(cont.)
Network Host
172.16.2.160
255.255.255.192
10101100
11111111
10101100
00010000
11111111
00010000
11111111
00000010
10100000
11000000
10000000
00000010
Subnet
•Network number extended by ten bits
16172 2 128
Network
Number
128
192
224
240
248
252
254
255
128
192
224
240
248
252
254
255

147
Subnet Mask Exercise
Address Subnet Mask Class Subnet
172.16.2.10
10.6.24.20
10.30.36.12
255.255.255.0
255.255.240.0
255.255.255.0

148
Subnet Mask Exercise Answers
Address Subnet Mask Class Subnet
172.16.2.10
10.6.24.20
10.30.36.12
255.255.255.0
255.255.240.0
255.255.255.0
B
A
A
172.16.2.0
10.6.16.0
10.30.36.0

149
Broadcast Addresses
172.16.1.0
172.16.2.0
172.16.3.0
172.16.4.0
172.16.3.255
(Directed Broadcast)
255.255.255.255
(Local Network Broadcast)
X
172.16.255.255
(All Subnets Broadcast)

150
Addressing Summary Example
10101100
11111111
10101100
00010000
11111111
00010000
11111111
00000010
10100000
11000000
10000000
00000010
10101100 00010000 00000010 10111111
10101100 00010000 00000010 10000001
10101100 00010000 00000010 10111110
Host
Mask
Subnet
Broadcast
Last
First
172.16.2.160
255.255.255.192
172.16.2.128
172.16.2.191
172.16.2.129
172.16.2.190
1
2
3
4
5
6
7
89
16172 2 160

151
IP Host Address: 172.16.2.121
Subnet Mask: 255.255.255.0
• Subnet Address = 172.16.2.0
• Host Addresses = 172.16.2.1–172.16.2.254
• Broadcast Address = 172.16.2.255
• Eight Bits of Subnetting
Network Subnet Host
10101100 00010000 00000010 11111111
172.16.2.121:
255.255.255.0:
10101100
11111111
Subnet: 10101100 00010000
00010000
11111111
00000010
00000010
11111111
01111001
00000000
00000000
Class B Subnet Example
Broadcast:
Network

152
Subnet Planning
Other
Subnets
192.168.5.16
192.168.5.32 192.168.5.48
20 Subnets
5 Hosts per Subnet
Class C Address:
192.168.5.0

153
11111000
IP Host Address: 192.168.5.121
Subnet Mask: 255.255.255.248
Network Subnet Host
192.168.5.121: 11000000
11111111
Subnet: 11000000 10101000
10101000
11111111
00000101
00000101
11111111
01111001
01111000
255.255.255.248:
Class C Subnet Planning
Example
• Subnet Address = 192.168.5.120
• Host Addresses = 192.168.5.121–192.168.5.126
• Broadcast Address = 192.168.5.127
• Five Bits of Subnetting
Broadcast:
NetworkNetwork
11000000 10101000 00000101 01111111

154
Exercise
• 192.168.10.0
• /27
? – SNM
? – Block Size
?- Subnets

155
Exercise
• /27
? – SNM – 224
? – Block Size = 256-224 = 32
?- Subnets
Subnets 10.0 10.32 10.64
FHID 10.1 10.33
LHID 10.30 10.62
Broadcast 10.31 10.63

156
Exercise
• 192.168.10.0
• /30
? – SNM
? – Block Size
?- Subnets

157
Exercise
• /30
? – SNM – 252
? – Block Size = 256-252 = 4
?- Subnets
Subnets 10.0 10.4 10.8
FHID 10.1 10.5
LHID 10.2 10.6
Broadcast 10.3 10.7

158
Exercise
Mask Subnets Host
/26 ? ? ?
/27 ? ? ?
/28 ? ? ?
/29 ? ? ?
/30 ? ? ?

159
Exercise
Mask Subnets Host
/26 192 4 62
/27 224 8 30
/28 240 16 14
/29 248 32 6
/30 252 64 2

160
Exam Question
• Find Subnet and Broadcast address
– 192.168.0.100/27

161
Exercise
192.168.10.54 /29
Mask ?
Subnet ?
Broadcast ?

162
Exercise
192.168.10.130 /28
Mask ?
Subnet ?
Broadcast ?

163
Exercise
192.168.10.193 /30
Mask ?
Subnet ?
Broadcast ?

164
Exercise
192.168.1.100 /26
Mask ?
Subnet ?
Broadcast ?

165
Exercise
192.168.20.158 /27
Mask ?
Subnet ?
Broadcast ?

166
Class B
172.16.0.0 /19
Subnets ?
Hosts ?
Block Size ?

167
Class B
172.16.0.0 /19
Subnets 23 -2 = 6
Hosts 213 -2 = 8190
Block Size 256-224 = 32
Subnets 0.0 32.0 64.0 96.0
FHID 0.1 32.1 64.1 96.1
LHID 31.254 63.254 95.254 127.254
Broadcast 31.255 63.255 95.255 127.255

168
Class B
172.16.0.0 /27
Subnets ?
Hosts ?
Block Size ?

169
Class B
172.16.0.0 /27
Subnets 211 -2 = 2046
Hosts 25 -2 = 30
Block Size 256-224 = 32
Subnets 0.0 0.32 0.64 0.96
FHID 0.1 0.33 0.65 0.97
LHID 0.30 0.62 0.94 0.126
Broadcast 0.31 0.63 0.95 0.127

170
Class B
172.16.0.0 /23
Subnets ?
Hosts ?
Block Size ?

171
Class B
172.16.0.0 /23
Subnets 27 -2 = 126
Hosts 29 -2 = 510
Block Size 256-254 = 2
Subnets 0.0 2.0 4.0 6.0
FHID 0.1 2.1 4.1 6.1
LHID 1.254 3.254 5.254 7.254
Broadcast 1.255 3.255 5.255 7.255

172
Class B
172.16.0.0 /24
Subnets ?
Hosts ?
Block Size ?

173
Class B
172.16.0.0 /24
Subnets 28 -2 = 254
Hosts 28 -2 = 254
Block Size 256-255 = 1
Subnets 0.0 1.0 2.0 3.0
FHID 0.1 1.1 2.1 3.1
LHID 0.254 1.254 2.254 3.254
Broadcast 0.255 1.255 2.255 3.255

174
Class B
172.16.0.0 /25
Subnets ?
Hosts ?
Block Size ?

175
Class B
172.16.0.0 /25
Subnets 29 -2 = 510
Hosts 27 -2 = 126
Block Size 256-128 = 128
Subnets 0.0 0.128 1.0 1.128 2.0 2.128
FHID 0.1 0.129 1.1 1.129 2.1 2.129
LHID 0.126 0.254 1.126 1.254 2.126 2.254
Broadcast 0.127 0.255 1.127 1.255 2.127 2.255

177
Find out Subnet and Broadcast
Address
• 172.16.85.30/29

178
Address
• 172.30.101.62/23

179
Address
• 172.20.210.80/24

180
Exercise
• Find out the mask which gives 100
subnets for class B

181
Exercise
• Find out the Mask which gives 100 hosts
for Class B

182
Class A
10.0.0.0 /10
Subnets ?
Hosts ?
Block Size ?

183
Class A
10.0.0.0 /10
Subnets 22 -2 = 2
Hosts 222 -2 = 4194302
Block Size 256-192 = 64
Subnets 10.0 10.64 10.128 10.192
FHID 10.0.0.1 10.64.0.1 10.128.0.1 10.192.0.1
LHID 10.63.255.254 10.127.255.254 10.191.255.254 10.254.255.254
Broadcast 10.63.255.255 10.127.255.255 10.191.255.255 10.254.255.255

184
Class A
10.0.0.0 /18
Subnets ?
Hosts ?
Block Size ?

185
Class A
10.0.0.0 /18
Subnets 210 -2 = 1022
Hosts 214 -2 = 16382
Block Size 256-192 = 64
Subnets 10.0.0.0 10.0.64.0 10.0.128.0 10.0.192.0
FHID 10.0.0.1 10.0.64.1 10.0.128.1 10.0.192.1
LHID 10.0.63.254 10.0.127.254 10.0.191.254 10.0.254.254
Broadcast 10.0.63.255 10.0.127.255 10.0.191.255 10.0.254.255

186
Broadcast Addresses Exercise
Address Class Subnet Broadcast
201.222.10.60 255.255.255.248
Subnet Mask
15.16.193.6 255.255.248.0
128.16.32.13 255.255.255.252
153.50.6.27 255.255.255.128

187
Broadcast Addresses Exercise
Answers
153.50.6.127
Address Class Subnet Broadcast
201.222.10.60 255.255.255.248 C 201.222.10.63201.222.10.56
Subnet Mask
15.16.193.6 255.255.248.0 A 15.16.199.25515.16.192.0
128.16.32.13 255.255.255.252 B 128.16.32.15128.16.32.12
153.50.6.27 255.255.255.128 B 153.50.6.0

188
VLSM
• VLSM is a method of designating a different subnet
mask for the same network number on different subnets
• Can use a long mask on networks with few hosts and a
shorter mask on subnets with many hosts
• With VLSMs we can have different subnet masks for
different subnets.

189
Variable Length Subnetting
 VLSM allows us to use one class C address to
design a networking scheme to meet the
following requirements:
Bangalore 60 Hosts
Mumbai 28 Hosts
Sydney 12 Hosts
Singapore 12 Hosts
WAN 1 2 Hosts
WAN 2 2 Hosts
WAN 3 2 Hosts

190
Networking Requirements
Bangalore 60
Mumbai 60 Sydney 60 Singapore 60
WAN 1
WAN 2
WAN 3
In the example above, a /26 was used to provide the 60 addresses
for Bangalore and the other LANs. There are no addresses left for
WAN links

191
Networking Scheme
Mumbai 192.168.10.64/27
Bangalore
192.168.10.0/26
Sydney 192.168.10.96/28
Singapore 192.168.10.112/28
WAN 192.168.10.129 and 130 WAN 192.198.10.133 and 134
WAN 192.198.10.137 and 138
60 12 12
28
2
2
2
192.168.10.128/30
192.168.10.136/30
192.168.10.132/30

192
VLSM Exercise
2
2
2
40
25
12
192.168.1.0

193
VLSM Exercise
2
2
2
40
25
12
192.168.1.0
192.168.1.4/30
192.168.1.8/30
192.168.1.12/30
192.168.1.16/28
192.168.1.32/27
192.168.1.64/26

194
VLSM Exercise
2
2
8
15
5
192.168.1.0
2
2
35

195
Summarization
• Summarization, also called route aggregation, allows
routing protocols to advertise many networks as one
address.
• The purpose of this is to reduce the size of routing tables
on routers to save memory
• Route summarization (also called route aggregation or
supernetting) can reduce the number of routes that a
router must maintain
• Route summarization is possible only when a proper
addressing plan is in place
• Route summarization is most effective within a
subnetted environment when the network addresses are
in contiguous blocks

197
Supernetting
Network Subnet
172.16.12.0 11000000
11111111
10101000
11111111
00001100
11111111255.255.255.0
NetworkNetwork
00000000
00000000
16 8 4 2 1
172.16.13.0 11000000 1010100000001101 00000000
172.16.14.0 11000000 10101000 00001110 00000000
172.16.15.0 11000000 10101000 00001111 00000000

198
Supernetting
Network Subnet
172.16.12.0 11000000
11111111
10101000
11111111
00001100
11111100255.255.252.0
NetworkNetwork
00000000
00000000
16 8 4 2 1
172.16.13.0 11000000 1010100000001101 00000000
172.16.14.0 11000000 10101000 00001110 00000000
172.16.15.0 11000000 10101000 00001111 00000000
172.16.12.0/24
172.16.13.0/24
172.16.14.0/24
172.16.15.0/24
172.16.12.0/22

199
Supernetting Question
 What is the most efficient summarization that TK1 can use to advertise its
networks to TK2?
A. 172.1.4.0/24172.1.5.0/24172.1.6.0/24172.1.7.0/24
B. 172.1.0.0/22
C. 172.1.4.0/25172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24
D. 172.1.0.0/21
E. 172.1.4.0/22

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